Solve laplace equation inside a rectangular 
My answer is $U = Acos(nπx/L)e^-nπy/L$
I really have no idea how to solve the particular solution. Please advise me.
 A: The separation of variables method leads to solutions of the form $\sin(\omega x)\sinh(\omega y)$ or $\sin(\omega x)\cosh(\omega y)$ or $\cos(\omega x)\sinh(\omega y)$ or $\cos(\omega x)\cosh(\omega y)$ or other equivalent forms with exp fonctions, where $\omega$ is any constant.
The conditions $\frac{\partial u}{\partial x}(0,y)=0$ and $\frac{\partial u}{\partial x}(L,y)=0$ draw us to select the solutions of the form $\cos(\omega x)\sinh(\omega y)$ or $\cos(\omega x)\cosh(\omega y)$ with $\omega=\frac{n\pi}{L}$ because the derivative of $\cos(\omega x)$ is $-\omega \sin(\omega x)$ which is equal to $0$ at $x=0$ and $x=L$.
The condition $u(x,0)=0$ draw us to select the form $\cos(\omega x)\sinh(\omega y)$ because $\sinh(\omega y)=0$ at $y=0$
So, at this stage, the general solution is :
$$u(x,y)=c_0+\Sigma_{n=1}^{\infty} c_n \cos(\frac{n\pi x}{L})\sinh(\frac{n\pi y}{L})$$
The condition $u(x,H)=f(x)$ is :
$$c_0+\Sigma_{n=1}^{\infty} c_n \cos(\frac{n\pi x}{L})\sinh(\frac{n\pi H}{L})=f(x)$$
The function $f(x)$ is expressed on the form of a Fourier series, taking account of the compatibility of $f(x)$ with the conditions at vertex $\frac{\partial u}{\partial x}(0,H)=0=f'(0)$ and $\frac{\partial u}{\partial x}(L,H)=0=f'(L)$:
$$f(x)=b_0+\Sigma_{n=1}^{\infty}b_n \cos(\frac{n\pi x}{L})$$
$f(x)$ is a given function, so the coefficients $b_n$ of the Fourier series are given, or can be computed from the definition of $f(x)$. Hense :
$c_0=b_0$ and $ c_n \sinh(\frac{n\pi H}{L})=b_n$ in $n>0$
The final result is :
$$u(x,y)=b_0+\Sigma_{n=1}^{\infty} \frac{b_n}{\sinh(\frac{n\pi H}{L})} \cos(\frac{n\pi x}{L})\sinh(\frac{n\pi y}{L})$$
In order to answer to the questions (i) ans (ii), you have to express the given functions $f(x)$ on the form of Fourier series and compute the corresponding coefficients $b_n$.
For example, a very simple case is :
(iii) $f(x)=\cos(\frac{m\pi x}{L})$
The Fourier series is reduced to one term only with $b_m$=1 and the result is :
$u(x,y)= \frac{1}{\sinh(\frac{m\pi H}{L})} \cos(\frac{m\pi x}{L})\sinh(\frac{m\pi y}{L})$
